1
$\begingroup$

For example, oxygen is neutral when it has just 6 electrons. But the pink O in the molecule pictured has 3 bonds giving it 8 electrons, a full-octet. It's more electronegative than all the atoms it's bonded to also. It would make sense if an O with a full-octet had a -2 charge since it has an extra 2 electrons.

But because it "owns" only 1 electron from each bond, the O has a positive charge. This doesn't make sense to me: if we don't count the extra electrons in its charge, why do we count them towards its octet? C2OH

$\endgroup$
10
  • $\begingroup$ Where do the bonding electrons come from? This question might help. $\endgroup$ Commented Sep 20, 2020 at 5:13
  • $\begingroup$ Umm, such issue winding up in sugar hydrolysis of all things is rather surprising... $\endgroup$
    – Mithoron
    Commented Sep 20, 2020 at 17:19
  • $\begingroup$ chemistry.stackexchange.com/questions/14478/… $\endgroup$
    – Mithoron
    Commented Sep 20, 2020 at 17:21
  • 1
    $\begingroup$ @histrionics no it's not. A formal charge is of course formal. The oxygen in your sketch has a real charge. Or the ensemble. But there is a charge somewhere. And it is more around that oxygen. $\endgroup$
    – Alchimista
    Commented Sep 22, 2020 at 18:36
  • 1
    $\begingroup$ @Mithoron I vote to reopen the question. The edited question asks about the difference in counting electrons when looking for octets as opposed to assigning formal charges. The linked question compares counting electrons for assigning oxidation states or formal charges. Ideally, there would be a question comparing all three ways of counting to direct all further questions there. $\endgroup$
    – Karsten
    Commented Sep 22, 2020 at 21:52

0